CF 111B Petya and Divisors

题目描述

Petya喜欢寻找数字的因子。有一天他看到了这样的一条题目$:$

有$n$个形如$[x,y]$的二元组，对于每一个二元组，Petya希望找到有多少个$x_i$的因子，使得在$[x_{i-y_i},x_{i-y_i+1},\cdots ,x_{i-1}]$中的每一个数都不能整除它。帮帮他解决这个问题吧。

输入输出格式

输入格式

第一行有一个正整数$n$，满足$(1\le n\le 10000)$

下面有$n$行，每行$2$个非负整数$x_i$与$y_i(1\le x_i\le 10^5,0\le y\le i-1)$，意义如上文所述。

特别的，如果$y_i=0$，结果就是$x_i$因子的个数。

输出格式

共$n$行，每行$1$个数，表示有多少个$k$，满足$[x_i $ $mod$ $k=0$并且$(\forall j:i-y_i \leq j < i) $ $mod$ $k\ne 0]$

说明

样例一中前五个数的符合条件的因子如下:

• $1)$ $1,2,4$
• $2)$ $3$
• $3)$ $5$
• $4)$ $2,6$
• $5)$ $9,18$

题目描述

Little Petya loves looking for numbers’ divisors. One day Petya came across the following problem:

You are given $ n $ queries in the form " $ x_{i} $ $ y_{i} $ ". For each query Petya should count how many divisors of number $ x_{i} $ divide none of the numbers $ x_{i-yi},x_{i-yi}+1,…,x_{i-1} $ . Help him.

输入格式

The first line contains an integer $ n $ ( $ 1<=n<=10^{5} $ ). Each of the following $ n $ lines contain two space-separated integers $ x_{i} $ and $ y_{i} $ ( $ 1<=x_{i}<=10^{5} $ , $ 0<=y_{i}<=i-1 $ , where $ i $ is the query’s ordinal number; the numeration starts with $ 1 $ ).

If $ y_{i}=0 $ for the query, then the answer to the query will be the number of divisors of the number $ x_{i} $ . In this case you do not need to take the previous numbers $ x $ into consideration.

输出格式

For each query print the answer on a single line: the number of positive integers $ k $ such that

样例 #1

样例输入 #1

6
4 0
3 1
5 2
6 2
18 4
10000 3
1
2
3
4
5
6
7

样例输出 #1

3
1
1
2
2
22
1
2
3
4
5
6

提示

Let’s write out the divisors that give answers for the first 5 queries:

1) 1, 2, 4

2) 3

3) 5

4) 2, 6

5) 9, 18

代码

#include

using namespace std;

int l[1000000];

int T;

int main()
{
	cin>>T;
	
	memset(l,-1,sizeof(l));
	
	for(int k=1;k<=T;k++)
	{		
		int x,y;
		
		int ans=0;
		
		cin>>x>>y;
		
		for(int i=1;i<=x/i;i++)
		{
			if(x%i==0)
			{
				if(k-l[i]>y)
				{
					ans++;
				}
				
				if(x-i*i!=0&&k-l[x/i]>y)
				{
					ans++;
				}
				
				l[i]=l[x/i]=k;
			}
		}
		
		cout<<ans<<endl;
	}	
	
	return 0;
} 
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